Answer
$1$
Work Step by Step
Recall:
(1) $\cot{(90^\circ-\theta)}=\tan{\theta}$
(2) $\csc{(90^\circ-\theta)}=\sec{\theta}$
Thus,
$\sec{35^\circ}\csc{55^\circ}-\tan{35^\circ}\cot{55^\circ}\\
=\sec{35^\circ}\sec{35^\circ}-\tan{35^\circ}\tan{35^\circ}\\
=\sec{35^\circ}\sec{(90^\circ-55^\circ)}-\tan{35^\circ}\tan{(90^\circ-55^\circ)}\\
=\sec^2{35^\circ}-\tan^2{35^\circ}$
Since $\sec^2{\theta}-\tan^2{\theta}=1$, then
$\sec^2{35^\circ}-\tan^2{35^\circ}=1$
